Chemistry, 22.02.2021 23:30 Uhmjujiooo4220

Stoichiometry: You conduct the following precipitation reaction in a lab:
CoCl₂ + 2NaOH → 2NaCl + Co(OH)₂
If you react 10.0 mL of 1.5 M CoCl₂ with plenty of NaOH, how many grams of Co(OH)₂ will precipitate out?

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Answer from: slonzyy8397

1.4 g Co(OH)₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Unit 0

Base 10 decimal system

Atomic Structure

Reading a Periodic TableMolarity = moles of solute / liters of solution

Aqueous Solutions

States of MatterPrediction Reactions RxN

Stoichiometry

Using Dimensional AnalysisAnalyzing Reactions RxNExplanation:

Step 1: Define

[RxN - Balanced] CoCl₂ + 2NaOH → 2NaCl + Co(OH)₂

[Given] 10.0 mL, 1.5 M CoCl₂

[Solve] grams Co(OH)₂

Step 2: Identify Conversions

[Base 10] 1000 mL = 1 L

[RxN] 1 mol CoCl₂ → 1 mol Co(OH)₂

[PT] Molar Mass of Co - 58.93 g/mol

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of Co(OH)₂ - 58.93 + 2(16.00) + 2(1.01) = 92.95 g/mol

Step 3: Stoich

[DA] Convert mL to L [Set up]: $\displaystyle 10.0 mL(\frac{1 \ L}{1000 \ mL})$ [DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.01 \ L$ [DA] Find moles of CoCl₂ [Molarity]: $\displaystyle 1.5 \ M \ CoCl_2 = \frac{x \ mol \ CoCl_2}{0.01 \ L}$ [DA] Solve for x [Multiplication Property of Equality]: $\displaystyle 0.015 \ mol \ CoCl_2$ [DA] Set up [Reaction Stoich]: $\displaystyle 0.015 \ mol \ CoCl_2(\frac{1 \ mol \ Co(OH)_2}{1 \ mol \ CoCl_2})(\frac{92.95 \ g \ Co(OH)_2}{1 \ mol \ Co(OH)_2})$ [DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.39425 \ g \ Co(OH)_2$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

1.39425 g Co(OH)₂ ≈ 1.4 g Co(OH)₂

Answer from: Quest

1 is the answer to this problem

Answer from: Quest

60.0 - there are 3 significant numbers

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Stoichiometry: You conduct the following precipitation reaction in a lab:
CoCl₂ + 2NaOH → 2Na...

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